Consider the real plane $\mathbb R^2$ and the shear mapping $$T(c) = \begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}$$
Is it possible to find an ellipse $E \equiv \frac{x^2}{a^2} + \frac{x^2}{b^2} =1$ whose image under $T_c$ is congruent to itself (the image of itself under an isometry)?
I know that the image of $E$ under $T_c$ is the subset having for equation
$$\frac{X^2}{a^2} + \frac{(Y-cX)^2}{b^2} =1$$
but I'm not able to select appropriate $a,b$ (depending on $c$) to get the conclusion or a contradiction.
The question is a subsequent question of this one.
Yes, it is possible. Take an ellipse passing through the points $A(0,1)$, $B(0,-1)$, $C(1,-c/2)$ and $D(-1,c/2)$, centered at the origin and with tangents at $C$ and $D$ being vertical.Your shear takes the given points to $T(A)=(0,1)$, $T(B)=(0,-1)$, $T(C)=(1,c/2)$ and $T(D)=(-1,-c/2)$ and your original ellipse to its mirror image about the $x$-axis. It is not possible to find an ellipse of the form you propose whose image is congruent to itself. You need a "rotated" one like the one I mentioned.